LeetCode Insert Interval

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

题意：合并一个区间。

思路：分情况看：首先假设对于new,end < cur.start，那么就非常easy的知道应该插入cur的前面：相反，此时假设new.start > cur.end的话。那么就能继续往下走了。否则就是这种情况：此时两个区间是有重叠的。各自取左右边界的最小和最大值。

package code;

import java.util.ArrayList;
import java.util.List;
import java.util.ListIterator;


public class Insert_Interval {
	
	public static void main(String[] args) {
		List<Interval> intervals = new ArrayList<Insert_Interval.Interval>();
		Insert_Interval root = new Insert_Interval();
		Insert_Interval.Interval s1 = root.new Interval(1, 3);
		intervals.add(s1);
		intervals.add(new Insert_Interval().new Interval(6, 9));
		Insert_Interval.Interval t = root.new Interval(2, 5);
		Insert_Interval.Solution solution = root.new Solution();
		List<Interval> ans = solution.insert(intervals, t);
		for (Interval interval : ans) {
			System.out.println(interval.start + "-" + interval.end);
		}
	}
	
	public class Interval {
		 int start;
		 int end;
		 Interval() { start = 0; end = 0; }
		 Interval(int s, int e) { start = s; end = e; }
	}
	
	public class Solution {
	    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
	    	if (intervals == null || newInterval == null)
	    		return intervals;
	    	if (intervals.size() == 0) 
	    		intervals.add(newInterval);
	    	
	    	ListIterator<Interval> it = intervals.listIterator();
	    	while (it.hasNext()) {
	    		Interval tmp = it.next();
	    		if (newInterval.end < tmp.start) {
	    			it.previous();
	    			it.add(newInterval);
	    			return intervals;
	    		} else {
	    			if (newInterval.start > tmp.end) 
	    				continue;
	    			else {
	    				newInterval.start = Math.min(tmp.start, newInterval.start);
	    				newInterval.end = Math.max(tmp.end, newInterval.end);
	    				it.remove();
	    			}
	    		}
	    	}
	    	
	    	intervals.add(newInterval);
	        return intervals;
	    }
	}
}

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